Note: The tickmarks on the horizontal axis are $$ -12 \pi,-8 \pi,-4 \pi, 4 \pi, 8 \pi, 12 \pi $$, respectively. Determine the period of the wave above. $$ \square $$ 8 Correct response: 2*4*Pi (1 Mark) Click on Verify to continue! (b) Assuming the Fourier series of $$ f(t) $$ is given by $$ f(t)=\frac{a_{0}}{2}+\sum_{n=1}^{\infty}\left(a_{n} \cos \frac{n \pi}{L} t+b_{n} \sin \frac{n \pi}{L} t\right) $$ where $$ L $$ is the half period. Complete the steps below: (i) Determine the Fourier coefficient $$ a_{0} $$. Enter EXACT value: $$ a_{0}=12 \pi $$ Correct response: $$ 12 \pi $$ (1 Mark) Click on Verify to continue! (ii) Determine the Fourier coefficient $$ a_{n} $$ NOTE: - In Symbolic mode the correct syntax is, say $$ \frac{-2}{n * \pi} *\left((-1)^{2}-3\right) $$. Take note of * sign. - In Text mode the correct syntax is $$ -2 /\left(n^{*} P i\right)^{*}\left((-1)^{\wedge} 2-3\right) $$. $$ a_{n}= $$

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UpStudy AI · Accepted Answer

The result could not be found for the integral. Let's simplify the expression before calculating the integral.

The expression for $$ a_{n} $$ is:

$$ a_{n} = \frac{1}{2} \int_{0}^{4 \pi} 12 \pi \cos \left( \frac{n}{4} t \right) dt $$

To simplify this expression, we can use the fact that the integral of a cosine function over a full period is zero. Since the period of the cosine function is $$ 2 \pi $$, the integral from $$ 0 $$ to $$ 4 \pi $$ can be broken down into four periods of the cosine function.

Therefore, the expression for $$ a_{n} $$ simplifies to:

$$ a_{n} = \frac{1}{2} \cdot 12 \pi \cdot 0 = 0 $$

So, the Fourier coefficient $$ a_{n} $$ is 0.
$$ a_{n} = 0 $$

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